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Should this code fail to compile in C++17?
What is a smart pointer and when should I use one?When should static_cast, dynamic_cast, const_cast and reinterpret_cast be used?How can I profile C++ code running on Linux?Why should I use a pointer rather than the object itself?Compiling an application for use in highly radioactive environmentsWhat are the new features in C++17?clang 3.9, auto_ptr and boostC++17 code not compiling on Travis with Clang-6.0C++17: explicit conversion function vs explicit constructor + implicit conversions - have the rules changed?C++17 - Binding rvalue reference to non-const lvalue ref
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I was updating a project to use C++17 and found a few instances where code that followed this pattern was causing a compile error on recent versions of clang:
#include <boost/variant.hpp>
struct vis : public boost::static_visitor<void>
void operator()(int) const
;
int main()
boost::variant<int> v = 0;
boost::apply_visitor(vis, v);
Using clang v8.0 in C++17 mode, this fails with the following error:
<source>:11:30: error: temporary of type 'boost::static_visitor<void>' has protected destructor
boost::apply_visitor(vis, v);
^
/opt/compiler-explorer/libs/boost_1_64_0/boost/variant/static_visitor.hpp:53:5: note: declared protected here
~static_visitor() = default;
However, it compiles cleanly in C++14 mode. I found that if I change the brace initialization vis to parentheses vis(), then it compiles correctly in both modes. Every version of gcc that I've tried allows both variants in C++17 mode.
Is this a correct change in behavior from C++14 to C++17, or is this a clang bug? If it is correct, why is it now invalid in C++17 (or maybe it always was, but clang just allows it in earlier standard revisions)?
c++ clang c++17 boost-variant
edited 8 hours ago
Barry
191k21345630
asked 8 hours ago
Jason RJason R
6,18223359
add a comment |
I was updating a project to use C++17 and found a few instances where code that followed this pattern was causing a compile error on recent versions of clang:
#include <boost/variant.hpp>
struct vis : public boost::static_visitor<void>
void operator()(int) const
;
int main()
boost::variant<int> v = 0;
boost::apply_visitor(vis, v);
Using clang v8.0 in C++17 mode, this fails with the following error:
<source>:11:30: error: temporary of type 'boost::static_visitor<void>' has protected destructor
boost::apply_visitor(vis, v);
^
/opt/compiler-explorer/libs/boost_1_64_0/boost/variant/static_visitor.hpp:53:5: note: declared protected here
~static_visitor() = default;
However, it compiles cleanly in C++14 mode. I found that if I change the brace initialization vis to parentheses vis(), then it compiles correctly in both modes. Every version of gcc that I've tried allows both variants in C++17 mode.
Is this a correct change in behavior from C++14 to C++17, or is this a clang bug? If it is correct, why is it now invalid in C++17 (or maybe it always was, but clang just allows it in earlier standard revisions)?
c++ clang c++17 boost-variant
edited 8 hours ago
Barry
191k21345630
asked 8 hours ago
Jason RJason R
6,18223359
It seem you use aggregate initialization instead of default constructor :-/ Changing tovis()compiles.
– Jarod42
8 hours ago
1
Boost fixed this in 1.70. It's also discussed here.
– interjay
8 hours ago
add a comment |
I was updating a project to use C++17 and found a few instances where code that followed this pattern was causing a compile error on recent versions of clang:
#include <boost/variant.hpp>
struct vis : public boost::static_visitor<void>
void operator()(int) const
;
int main()
boost::variant<int> v = 0;
boost::apply_visitor(vis, v);
Using clang v8.0 in C++17 mode, this fails with the following error:
<source>:11:30: error: temporary of type 'boost::static_visitor<void>' has protected destructor
boost::apply_visitor(vis, v);
^
/opt/compiler-explorer/libs/boost_1_64_0/boost/variant/static_visitor.hpp:53:5: note: declared protected here
~static_visitor() = default;
However, it compiles cleanly in C++14 mode. I found that if I change the brace initialization vis to parentheses vis(), then it compiles correctly in both modes. Every version of gcc that I've tried allows both variants in C++17 mode.
Is this a correct change in behavior from C++14 to C++17, or is this a clang bug? If it is correct, why is it now invalid in C++17 (or maybe it always was, but clang just allows it in earlier standard revisions)?
c++ clang c++17 boost-variant
edited 8 hours ago
Barry
191k21345630
asked 8 hours ago
Jason RJason R
6,18223359
I was updating a project to use C++17 and found a few instances where code that followed this pattern was causing a compile error on recent versions of clang:
#include <boost/variant.hpp>
struct vis : public boost::static_visitor<void>
void operator()(int) const
;
int main()
boost::variant<int> v = 0;
boost::apply_visitor(vis, v);
Using clang v8.0 in C++17 mode, this fails with the following error:
<source>:11:30: error: temporary of type 'boost::static_visitor<void>' has protected destructor
boost::apply_visitor(vis, v);
^
/opt/compiler-explorer/libs/boost_1_64_0/boost/variant/static_visitor.hpp:53:5: note: declared protected here
~static_visitor() = default;
However, it compiles cleanly in C++14 mode. I found that if I change the brace initialization vis to parentheses vis(), then it compiles correctly in both modes. Every version of gcc that I've tried allows both variants in C++17 mode.
Is this a correct change in behavior from C++14 to C++17, or is this a clang bug? If it is correct, why is it now invalid in C++17 (or maybe it always was, but clang just allows it in earlier standard revisions)?
c++ clang c++17 boost-variant
c++ clang c++17 boost-variant
edited 8 hours ago
Barry
191k21345630
asked 8 hours ago
Jason RJason R
6,18223359
edited 8 hours ago
Barry
191k21345630
asked 8 hours ago
Jason RJason R
6,18223359
edited 8 hours ago
Barry
191k21345630
edited 8 hours ago
Barry
191k21345630
edited 8 hours ago
Barry
191k21345630
191k21345630
asked 8 hours ago
Jason RJason R
6,18223359
asked 8 hours ago
Jason RJason R
6,18223359
asked 8 hours ago
Jason RJason R
6,18223359
6,18223359
It seem you use aggregate initialization instead of default constructor :-/ Changing tovis()compiles.
– Jarod42
8 hours ago
1
Boost fixed this in 1.70. It's also discussed here.
– interjay
8 hours ago
add a comment |
It seem you use aggregate initialization instead of default constructor :-/ Changing tovis()compiles.
– Jarod42
8 hours ago
1
Boost fixed this in 1.70. It's also discussed here.
– interjay
8 hours ago
It seem you use aggregate initialization instead of default constructor :-/ Changing to
vis() compiles.– Jarod42
8 hours ago
It seem you use aggregate initialization instead of default constructor :-/ Changing to
vis() compiles.– Jarod42
8 hours ago
1
1
Boost fixed this in 1.70. It's also discussed here.
– interjay
8 hours ago
Boost fixed this in 1.70. It's also discussed here.
– interjay
8 hours ago
add a comment |
1 Answer
1
active
oldest
votes
clang is correct here. Here's a reduced example:
struct B
protected:
B()
;
struct D : B ;
auto d = D;
In C++14, D is not an aggregate because it has a base class, so D is "normal" (non-aggregate) initialization which invokes D's default constructor, which in turn invokes B's default constructor. This is fine, because D has access to B's default constructor.
In C++17, the definition of aggregate was widened - base classes are now allowed (as long as they're non-virtual). D is now an aggregate, which means that D is aggregate initialization. And in aggregate-initialization, this means that we (the caller) are initializing all the subobjects - including the base class subobject. But we do not have access to B's constructor (it is protected), so we cannot invoke it, so it is ill-formed.
Fear not, the fix is easy. Use parentheses:
auto d = D();
This goes back to invoking D's default constructor as before.
answered 8 hours ago
BarryBarry
191k21345630
Perfect, that's what I was looking for: the widened definition of aggregate is the culprit.
– Jason R
8 hours ago
add a comment |
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
clang is correct here. Here's a reduced example:
struct B
protected:
B()
;
struct D : B ;
auto d = D;
In C++14, D is not an aggregate because it has a base class, so D is "normal" (non-aggregate) initialization which invokes D's default constructor, which in turn invokes B's default constructor. This is fine, because D has access to B's default constructor.
In C++17, the definition of aggregate was widened - base classes are now allowed (as long as they're non-virtual). D is now an aggregate, which means that D is aggregate initialization. And in aggregate-initialization, this means that we (the caller) are initializing all the subobjects - including the base class subobject. But we do not have access to B's constructor (it is protected), so we cannot invoke it, so it is ill-formed.
Fear not, the fix is easy. Use parentheses:
auto d = D();
This goes back to invoking D's default constructor as before.
answered 8 hours ago
BarryBarry
191k21345630
Perfect, that's what I was looking for: the widened definition of aggregate is the culprit.
– Jason R
8 hours ago
add a comment |
clang is correct here. Here's a reduced example:
struct B
protected:
B()
;
struct D : B ;
auto d = D;
In C++14, D is not an aggregate because it has a base class, so D is "normal" (non-aggregate) initialization which invokes D's default constructor, which in turn invokes B's default constructor. This is fine, because D has access to B's default constructor.
In C++17, the definition of aggregate was widened - base classes are now allowed (as long as they're non-virtual). D is now an aggregate, which means that D is aggregate initialization. And in aggregate-initialization, this means that we (the caller) are initializing all the subobjects - including the base class subobject. But we do not have access to B's constructor (it is protected), so we cannot invoke it, so it is ill-formed.
Fear not, the fix is easy. Use parentheses:
auto d = D();
This goes back to invoking D's default constructor as before.
answered 8 hours ago
BarryBarry
191k21345630
Perfect, that's what I was looking for: the widened definition of aggregate is the culprit.
– Jason R
8 hours ago
add a comment |
clang is correct here. Here's a reduced example:
struct B
protected:
B()
;
struct D : B ;
auto d = D;
In C++14, D is not an aggregate because it has a base class, so D is "normal" (non-aggregate) initialization which invokes D's default constructor, which in turn invokes B's default constructor. This is fine, because D has access to B's default constructor.
In C++17, the definition of aggregate was widened - base classes are now allowed (as long as they're non-virtual). D is now an aggregate, which means that D is aggregate initialization. And in aggregate-initialization, this means that we (the caller) are initializing all the subobjects - including the base class subobject. But we do not have access to B's constructor (it is protected), so we cannot invoke it, so it is ill-formed.
Fear not, the fix is easy. Use parentheses:
auto d = D();
This goes back to invoking D's default constructor as before.
answered 8 hours ago
BarryBarry
191k21345630
clang is correct here. Here's a reduced example:
struct B
protected:
B()
;
struct D : B ;
auto d = D;
In C++14, D is not an aggregate because it has a base class, so D is "normal" (non-aggregate) initialization which invokes D's default constructor, which in turn invokes B's default constructor. This is fine, because D has access to B's default constructor.
In C++17, the definition of aggregate was widened - base classes are now allowed (as long as they're non-virtual). D is now an aggregate, which means that D is aggregate initialization. And in aggregate-initialization, this means that we (the caller) are initializing all the subobjects - including the base class subobject. But we do not have access to B's constructor (it is protected), so we cannot invoke it, so it is ill-formed.
Fear not, the fix is easy. Use parentheses:
auto d = D();
This goes back to invoking D's default constructor as before.
answered 8 hours ago
BarryBarry
191k21345630
answered 8 hours ago
BarryBarry
191k21345630
answered 8 hours ago
BarryBarry
191k21345630
answered 8 hours ago
BarryBarry
191k21345630
191k21345630
Perfect, that's what I was looking for: the widened definition of aggregate is the culprit.
– Jason R
8 hours ago
add a comment |
Perfect, that's what I was looking for: the widened definition of aggregate is the culprit.
– Jason R
8 hours ago
Perfect, that's what I was looking for: the widened definition of aggregate is the culprit.
– Jason R
8 hours ago
Perfect, that's what I was looking for: the widened definition of aggregate is the culprit.
– Jason R
8 hours ago
add a comment |
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It seem you use aggregate initialization instead of default constructor :-/ Changing to
vis()compiles.– Jarod42
8 hours ago
1
Boost fixed this in 1.70. It's also discussed here.
– interjay
8 hours ago